College

You are helping with some repairs at home. You drop a hammer, and it hits the floor at a speed of 4 feet per second. If the acceleration due to gravity [tex](g)[/tex] is 32 feet/second [tex]^{2}[/tex], how far above the ground [tex](h)[/tex] was the hammer when you dropped it?

Use the formula:
[tex]v = \sqrt{2gh}[/tex]

A. 0.5 feet
B. 0.25 feet
C. 1.0 foot
D. 16.0 feet

Answer :

To solve this problem, we want to find out how far above the ground the hammer was when it was dropped. We use the formula:

[tex]\[ v = \sqrt{2gh} \][/tex]

where:
- [tex]\( v \)[/tex] is the speed of the hammer when it hits the floor (4 feet per second),
- [tex]\( g \)[/tex] is the acceleration due to gravity (32 feet/second²),
- [tex]\( h \)[/tex] is the height above the ground from which the hammer was dropped.

We need to solve for [tex]\( h \)[/tex]. Here's how you do it step-by-step:

1. Square Both Sides to Eliminate the Square Root:

Start with the formula:
[tex]\[ v = \sqrt{2gh} \][/tex]

Square both sides:
[tex]\[ v^2 = 2gh \][/tex]

2. Rearrange the Equation to Solve for [tex]\( h \)[/tex]:

Divide both sides by [tex]\( 2g \)[/tex] to isolate [tex]\( h \)[/tex]:
[tex]\[ h = \frac{v^2}{2g} \][/tex]

3. Substitute the Given Values:

Plug in the values for [tex]\( v \)[/tex] and [tex]\( g \)[/tex]:
[tex]\[ h = \frac{4^2}{2 \times 32} \][/tex]

4. Calculate [tex]\( h \)[/tex]:

Simplify the equation:
- [tex]\( v^2 = 4^2 = 16 \)[/tex]
- [tex]\( 2 \times 32 = 64 \)[/tex]

Now compute:
[tex]\[ h = \frac{16}{64} \][/tex]

5. Simplify the Fraction:

[tex]\[ h = \frac{1}{4} = 0.25 \][/tex]

So, the hammer was dropped from a height of 0.25 feet above the ground. The correct answer is:

B. 0.25 feet